تحليل: Volumetric Analysis PDF

‫اﻟﺗﺣﻠﯾل ‪Analysis‬‬ ‫ﺗﺣﻠﯾل طﯾﻔﻲ ﺿوﺋﻲ‬ ‫ﺗﺣﻠﯾل ﻛﮭرﻛﯾﻣﯾﺎﺋﻲ ‪ ،‬ﺗﺣﻠﯾل ﺑطراﺋﻖ اﻟﻛروﻣﺎﺗوﻏراﻓﯾﺎ‬ ‫د‪.‬ﻋﻼ ﯾوﻧس‬ ‫وﺟﻣﯾﻊ طراﺋﻖ اﻟﺗﺣﻠﯾل اﻵﻟﻲ‬ ‫اﻟﺗﺣﻠﯾل اﻟﺣﺟﻣﻲ ‪volumetric analysis‬‬ ‫د‪.‬ﻋﻼ ﯾوﻧس‬  ‫أﺳﺎﺳﯾﺎت‬...

‫اﻟﺗﺣﻠﯾل ‪Analysis‬‬ ‫ﺗﺣﻠﯾل طﯾﻔﻲ ﺿوﺋﻲ‬ ‫ﺗﺣﻠﯾل ﻛﮭرﻛﯾﻣﯾﺎﺋﻲ ‪ ،‬ﺗﺣﻠﯾل ﺑطراﺋﻖ اﻟﻛروﻣﺎﺗوﻏراﻓﯾﺎ‬ ‫د‪.‬ﻋﻼ ﯾوﻧس‬ ‫وﺟﻣﯾﻊ طراﺋﻖ اﻟﺗﺣﻠﯾل اﻵﻟﻲ‬ ‫اﻟﺗﺣﻠﯾل اﻟﺣﺟﻣﻲ ‪volumetric analysis‬‬ ‫د‪.‬ﻋﻼ ﯾوﻧس‬  ‫أﺳﺎﺳﯾﺎت‬ stoichiometry: The quantitative study of reactants and products in a chemical reaction. ‫اﻟﻧﺳب اﻟﻣوﻟﯾﺔ ﺑﯾن اﻟﻣواد ﻓﻲ اﻟﺗﻔﺎﻋل اﻟﻛﯾﻣﯾﺎﺋﻲ اﻟﻣﺗوازن‬ Stoichiometric amounts: The exact molar amounts of reactants and products that appear in the balanced chemical equation. Volumetric titrimetry: methods that require that a reagent solution of known concentration, standard solution or titrant, be used ‫ ﻋﻼ ﯾوﻧس‬.‫د‬ ‫د‪.‬ﻋﻼ ﯾوﻧس‬  ‫طرق اﻟﺗﻌﺑﯾر ﻋن اﻟﺗرﻛﯾز‪:‬‬ ‫‪g/L -1‬‬ ‫ﯾﻣﻛن اﻟﺗﻌﺑﯾرﻋن اﻟﺗرﻛﯾز ‪ concentration‬ﺑﻌدد اﻟﻐراﻣﺎت)‪g‬‬ ‫( اﻟﻣذاﺑﺔ ﻓــﻲ ﺣﺟم اﻟﻣﺣﻠول ﺑﺎﻟﻠﺗر)‪ ،V(L‬أو اﻟﻣﻠﯾﻐرام ﻟﻛل‬ ‫ﻣﻠﯾﻠﺗر )‪ (mg/mL‬وھﻧﺎك وﺣدات اﺧرى ﻣﺷﺗﻘﺔ ﺗﺳﺗﺧدم‬ ‫ﻟﻠﺗﻌﺑﯾرﻋن اﻟﺗرﻛﯾز وھﻲ اﻟﻣﻠﯾﻐرام ﻟﻛل ﻟﺗر )‪ (mg/L‬وﺗﻌرف‬ ‫اﺧﺗﺻﺎرا ﺑـ ‪(ppm) part per million‬ﺟزء ﻣن ﻣﻠﯾون‬ ‫ﺟزء‪ ،‬وﻛذﻟك اﻟﻣﯾﻛروﻏرام ﻟﻛل ﻟﺗر )‪ (µg/L‬وﺗﻌرف ﺑـ )‪(ppb‬‬ ‫‪ part per billion‬ﺟزء ﻣن ﺑﻠﯾون ﺟزء‪.‬‬ ‫د‪.‬ﻋﻼ ﯾوﻧس‬  ‫‪ - 2‬طرق اﻟﻧﺳﺑﺔ اﻟﻣﺋوﯾﺔ‬ ‫اﻟﻧﺳﺑﺔ اﻟﻣﺋوﯾﺔ اﻟوزﻧﯾﺔ ‪(W/W) %‬‬ ‫ ‬ ‫وﯾﻌﺑر ﻋﻧﮭﺎ ﺑوزن اﻟﻣذاب ﺑﺎﻟﻐراﻣﺎت اﻟﻣوﺟودة ﻓﻲ ‪100g‬‬ ‫ ‬ ‫ﻣن اﻟﻣﺣﻠول‬ ‫‪ × 100‬وزن اﻟﻣذاب ﺑﺎﻟﻐرام =‪%W/W‬‬ ‫وزن اﻟﻣﺣﻠول ﺑﺎﻟﻐرام‬ ‫اﻟﻧﺳﺑﺔ اﻟوزﻧﯾﺔ اﻟﺣﺟﻣﯾﺔ ‪(w/v)%‬‬ ‫ ‬ ‫وﯾﻌﺑر ﻋﻧﮭﺎ ﺑوزن اﻟﻣذاب ﺑﺎﻟﻐراﻣﺎت اﻟﻣوﺟودة ﻓﻲ ‪ 100‬ﻣل‬ ‫ ‬ ‫ﻣن اﻟﻣﺣﻠول‬ ‫د‪.‬ﻋﻼ ﯾوﻧس‬  ‫ اﻟﻧﺳﺑﺔ اﻟﻣﺋوﯾﺔ اﻟﺣﺟﻣﯾﺔ ‪(V/V)%‬‬ ‫ﯾﻌﺑر ﻋﻧﮭﺎ ﺑﻌدد ﻣﻠﯾﻠﺗرات اﻟﻣﺎدة اﻟﻣذاﺑﺔ‬ ‫واﻟﻣوﺟودة ﻓﻲ ‪100mL‬ﻣن اﻟﻣﺣﻠول‬ ‫ﺣﺟم اﻟﻣذاب ﺑﺎﻟﻣﻠﻠﺗر‬ ‫=‪(V/V)%‬‬ ‫‪× 100‬‬ ‫ﺣﺟم اﻟﻣﺣﻠول ﺑﺎﻟﻣﻠﻠﺗر‬ ‫د‪.‬ﻋﻼ ﯾوﻧس‬ ‫)‪Molarity (M‬‬ ‫‪ -3‬اﻟﻣوﻻرﯾﺔ‬ ‫د‪.‬ﻋﻼ ﯾوﻧس‬  ‫ﯾﻣﻛن ﻣن اﻟﻌﻼﻗﺎت اﻵﺗﯾﺔ ﺗﻌﯾﯾن ﻋدد اﻟﻣوﻻت)‪ (moles‬او‬ ‫اﻟﻣﻠﯾﻣوﻻت)‪ (mmoles‬ﻓﻲ ﺣﺟم ﻣﻌﯾن ﻣن اﻟﻣﺣﻠول‬ ‫‪ moles=M × VL‬ﻋدد اﻟﻣوﻻت‬ ‫‪ mmoles=M × VmL‬ﻋدداﻟﻣﻠﯾﻣوﻻت‬ ‫ﺣﯾث ‪ VL‬ﺣﺟم اﻟﻣﺣﻠول ﺑﺎﻟﻠﺗر و ‪ VmL‬ﺣﺟم اﻟﻣﺣﻠول ﺑﺎﻟﻣﻠﯾﻠﺗر‬ ‫د‪.‬ﻋﻼ ﯾوﻧس‬  ‫اﻟوزن اﻟﻣﻛﺎﻓﺊ‬ ‫)‪Equivalent weight(Eq‬‬ ‫ اﻟوزن اﻟﻣﻛﺎﻓﺊ ﻟﻠﺣﻣض‬ ‫ اﻟوزن اﻟﻣﻛﺎﻓﺊ ﻟﻸﺳﺎس‬ ‫ اﻟوزن اﻟﻣﻛﺎﻓﺊ ﻟﻠﻣﻠﺢ‪ :‬ھو اﻟوزن اﻟذي ﯾﺗﻔﺎﻋل ﻣﻊ وزن ﻣﻛﺎﻓﺊ ﻣن‬ ‫ﺣﺎﻣض أو أﺳﺎس‪.‬‬ ‫أﻣﺛﻠﺔ‪:‬‬ ‫ﻋﻧد ﺣﺳﺎب اﻟوزن اﻟوزن اﻟﻣﻛﺎﻓﺊ ﻟﻛرﺑوﻧﺎت اﻟﺻودﯾوم ﻻﺑد ﻣن‬ ‫اﻹﺷﺎرة ﻟﻠﺗﻔﺎﻋل اﻟذي ﺗدﺧل ﻓﯾﮫ ﻓﻣـﺛﻼَ‬ ‫ﻓﻲ اﻟﺗﻔﺎﻋل اﻟﺗﺎﻟﻲ‪:‬‬ ‫‪Na2CO3 + HCl → NaHCO3 + NaCl‬‬ ‫اﻟوزن اﻟﻣﻛﺎﻓﺊ = اﻟوزن اﻟﺟزﯾﺋﻲ‬ ‫د‪.‬ﻋﻼ ﯾوﻧس‬  ‫ ﺑﯾﻧﻣﺎ ﻓﻲ اﻟﺗﻔﺎﻋل اﻟﺗﺎﻟﻲ‪:‬‬ ‫‪Na2CO3 + 2HCl → NaCl + H2O + CO2‬‬ ‫اﻟوزن اﻟﻣﻛﺎﻓﺊ = اﻟوزن اﻟﺟزﯾﺋﻲ ‪2 /‬‬ ‫د‪.‬ﻋﻼ ﯾوﻧس‬  ‫)‪Normality ( N‬‬ ‫‪ - 4‬اﻟﻧظﺎﻣﯾﺔ‬ ‫ وﺗﻣﺛل ﻋدد اﻟﻣﻛﺎﻓﺋﺎت اﻟﻐراﻣﯾﺔ ﻣن اﻟﻣﺎدة اﻟﻣذاﺑﺔ ﻓﻲ ﻟﺗر ﻣن‬ ‫اﻟﻣﺣﻠول‪ ،‬واﻟﻣﺣﻠول اﻟﻧظﺎﻣﻲ ﯾﻣﻛن اﻋﺗﺑﺎره ذﻟك اﻟﻣﺣﻠول اﻟذي‬ ‫ﯾﺣﺗوي ﻋﻠﻰ اﻟوزن اﻟﻣﻛﺎﻓﺊ اﻟﻐراﻣﻲ ﻣن اﻟﻣﺎدة ﻓﻲ ﻟﺗر ﻣن‬ ‫اﻟﻣﺣﻠول‪ ،‬ﻓﻌﻧد اذاﺑﺔ ‪ 40g‬ﻣن ھﯾدروﻛﺳﯾد اﻟﺻودﯾوم ﻓﻲ‬ ‫ﻛﻣﯾﺔ ﻣن اﻟﻣﺎء وﯾﻛﻣل اﻟﺣﺟم اﻟﻰ ﻟﺗر ﻓﺎن ھذا اﻟﻣﺣﻠول‬ ‫ﻧظﺎﻣﯾﺗﮫ ﺗﺳﺎوي اﻟواﺣد‬ ‫‪(C g/l ) = N. Eq‬‬ ‫د‪.‬ﻋﻼ ﯾوﻧس‬  ‫ ﯾﻣﻛن ﻣن اﻟﻌﻼﻗﺎت اﻵﺗﯾﺔ ﺗﻌﯾﯾن ﻋدد اﻟﻣﻛﺎﻓﺋﺎت اﻟﻐراﻣﯾﺔ) ‪g.‬‬ ‫‪ (eq.‬او اﻟﻣﻠﯾﻐراﻣﺎت اﻟﻣﻛﺎﻓﺋﺔ)‪ (mg.eq.‬ﻓﻲ ﺣﺟم ﻣﻌﯾن ﻣن‬ ‫اﻟﻣﺣﻠول ﺑﻣﻌرﻓﺔ اﻟﻧظﺎﻣﯾﺔ‬ ‫ ‪ g.eq.=N × VL‬ﻋدد اﻟﻣﻛﺎﻓﺋﺎت اﻟﻐراﻣﯾﺔ‬ ‫ ‪ mg.eq.=N × Vml‬ﻋدد اﻟﻣﯾﻠﻲ ﻣﻛﺎﻓﺋﺎت اﻟﻐراﻣﯾﺔ‬ ‫د‪.‬ﻋﻼ ﯾوﻧس‬  ‫اﻟوزن اﻟﻣﻛﺎﻓﺊ ﻷي ﻣﺎدة ﺗدﺧل ﻓﻲ ﺗﻔﺎﻋﻼت اﻷﻛﺳدة‬ ‫ ‬ ‫واﻹرﺟﺎع ھو اﻟوزن اﻟﺟزﯾﺋﻲ ﻋﻠﻰ ﻋدد اﻻﻟﻛﺗروﻧﺎت‬ ‫اﻟﻣﻛﺗﺳﺑﺔ أو اﻟﻣﻔﻘودة‬ ‫اﺣﺳب اﻟوزن اﻟﻣﻛﺎﻓﺊ ﻟﻣﺎﯾﻠﻲ‬ ‫ ‬ ‫‪ -1‬ﻛﺑرﯾﺗﺎت اﻟﺣدﯾد‬ ‫ ‬ ‫‪ H2C2O4 -2‬ﺣﻣض اﻟﺣﻣﺎض‬ ‫ ‬ ‫‪ -3‬ﺑرﻣﻧﻐﻧﺎت اﻟﺑوﺗﺎﺳﯾوم ‪ KMnO4‬ﻓﻲ وﺳط ﺣﻣﺿﻲ‬ ‫ ‬ ‫‪ -4‬اﻟﯾود ‪I2‬‬ ‫ ‬ ‫ﺣﻣض اﻟﻛﺑرﯾت‬ ‫‪ -5‬ھﯾدروﻛﺳﯾد اﻟﻛﺎﻟﺳﯾوم‪ ،‬د‪.‬ﻋﻼ ﯾوﻧس‬ ‫ ‬  ‫ﺗﻣدﯾد اﻟﻣﺣﺎﻟﯾل‪:‬‬ ‫ ﯾﻣﻛن ﺣﺳﺎب ﻛﻣﯾﺔ اﻟﻣﺎدة اﻟﻣذاﺑﺔ ﻓﻲ ﻣﺣﻠول ذي ﺗرﻛﯾز ﻣﻌﯾن‬ ‫ﻣن اﻟﻌﻼﻗﺔ اﻵﺗﯾﺔ‪:‬‬ ‫ ﻛﻣﯾﺔ اﻟﻣﺎدة اﻟﻣذاﺑﺔ = اﻟﺣﺟم × اﻟﺗرﻛﯾز‬ ‫ وﻋﻧد ﺗﺧﻔﯾف اﻟﻣﺣﻠول ﻓﺎن اﻟﺣﺟم ﯾزداد وﻟﻛن اﻟﺗرﻛﯾز ﯾﻘل‬ ‫ﺑﯾﻧﻣﺎ ﺗظل ﻛﻣﯾﺔ اﻟﻣﺎدة ﺛﺎﺑﺗﺔ ‪ ،‬واﻟﻌﻼﻗﺔ اﻵﺗﯾﺔ ﺗرﺑط ﺑﯾن ﺣﺟم‬ ‫اﻟﻣﺣﻠول اﻟﻣرﻛز)‪ ( V1‬وﺗرﻛﯾزه ) ‪ ( C1‬وﺣﺟم اﻟﻣﺣﻠول‬ ‫اﻟﻣﺧﻔف ) ‪ ( V2‬وﺗرﻛﯾزه ) ‪.( C2‬‬ ‫ ‪C1 X V1 =C2 X V2‬‬ ‫د‪.‬ﻋﻼ ﯾوﻧس‬  ‫طرق ﺗﺣﺿﯾر ﺗراﻛﯾز ﻣﺧﺗﻠﻔﺔ ﻣن اﻷﺣﻣﺎض‬ ‫واﻟﻘواﻋد اﻟﺳﺎﺋﻠﺔ‪:‬‬ ‫ ﯾﺗﺿﻣن اﻟوﻋﺎء اﻟذي ﯾﺣﺗوي ﻋﻠﻰ اﻟﻣﺎدة ﻣﻌﻠوﻣﺎت ﻣﺛل اﻟﻧﺳﺑﺔ اﻟﻣﺋوﯾﺔ‬ ‫ﻟﻠﻣﺎدة )‪ ( %‬واﻟﻛﺛﺎﻓﺔ)‪ (d‬واﻟوزن اﻟﺟزﯾﺋﻲ)‪....،(M.wt‬وﻣن ﺧﻼل ھذه‬ ‫اﻟﻣﻌﻠوﻣﺎت ﯾﻣﻛﻧﻧﺎ أن ﻧﺣﺻل ﻋﻠﻰ اﻟﻣوﻻرﯾﺔ )‪ ( M‬واﻟﻧظﺎﻣﯾﺔ )‪( N‬‬ ‫ﻟﻠﻣﺎدة اﻟﺳﺎﺋﻠﺔ‪.‬‬ ‫‪- Density is the weight per unit volume at the specified‬‬ ‫‪temperature, usually g/ml or g/ cm3‬‬ ‫‪at 20 °C‬‬ ‫‪- Specific gravity is the ratio of the densities of the‬‬ ‫‪mass of a body (e.g, a solution), usually at 20 °C, to‬‬ ‫‪the mass of an equal volume‬‬ ‫د‪.‬ﻋﻼ ﯾوﻧس‬ ‫‪of water at 4 °C.‬‬  Describe the preparation of 500 ml of 0.100M sulfuric acid solution from the concentrated sulfuric acid, 94% density 1.831g/ml ‫ ﻋﻼ ﯾوﻧس‬.‫د‬ What is the molarity of K+ in a solution that contains 50 ppm of K3Fe(CN)6 (fw= 329.3)? Calculate the concentration of sodium ion in grams per liter after mixing 100 ml of 0.250 M NaCL and 200 ml of 0.1 M Na2SO4 ‫ ﻋﻼ ﯾوﻧس‬.‫د‬ -A solution is prepared by dissolving 1.5 g AgNO3 in a 500 ml volumetric flask and diluting to volume with water. Calculate the molarity of silver nitrate solution. How many millimoles AgNO3 were dissolved in the volumetric flask? ‫ ﻋﻼ ﯾوﻧس‬.‫د‬ ‫ﻛﯾف ﺗﺣﺿر ﻣﺣﻠول ﻗﯾﺎﺳﻲ ﺗرﻛﯾزه ‪ N 0.1‬ﻣن ﺣﻣض اﻟﻔوﺳﻔور ﻓﻲ ﺣﺟم‬ ‫ﻣﻘداره ‪ 250‬ﻣل ﻋﻠﻣﺎ ً أن ﻛﺛﺎﻓﺔ اﻟﺣﻣض ‪ 1.69‬غ‪ /‬ﻣل وﻧﺳﺑﺗﮫ اﻟﻣﺋوﯾﺔ‬ ‫‪%85‬‬ ‫د‪.‬ﻋﻼ ﯾوﻧس‬ ‫ﻟﺘﺨﻔﯿﻒ ﺗﺮﻛﯿﺰ ﻣﺤﻠﻮل إﻣﺎ ﻧﻤﺪد ﻛﻤﺎ ﻣﺮ ﺳﺎﺑﻘﺎ ً أو ﻧﻀﯿﻒ ﻣﺤﻠﻮل‬ :‫ذي ﺗﺮﻛﯿﺰ أﻗﻞ ﻣﺜﺎل‬ What volume of 1 M HCl solution must be added to 10 ml of 9 M HCl solution to give a solution 5 M of HCl. ‫ ﻋﻼ ﯾوﻧس‬.‫د‬  ‫ﻟﺰﯾﺎدة ﺗﺮﻛﯿﺰ ﻣﺤﻠﻮل ﻧﻘﻮم ﺑﺎﺳﺘﺨﺪام ﻣﺤﻠﻮل ذي ﺗﺮﻛﯿﺰ‬ ‫أﻛﺒﺮ أو ﻧﻀﯿﻒ ﻣﺎدة ﺻﻠﺒﺔ‬ ‫ﻣﺜﺎل‬ ‫‪What volume of 0.40 M Ba(OH)2 solution must be‬‬ ‫‪added to 50 ml of 0.30 M NaOH to give a solution‬‬ ‫‪0.50 M in OH-‬‬ ‫د‪.‬ﻋﻼ ﯾوﻧس‬ A 2 liters of 5300 ppm Na2CO3 Solution must be prepared by adding an amount of solid Na2CO3 to a 0.08N solution. How many grams of Na2CO3 must be added to obtain this solution. ‫ ﻋﻼ ﯾوﻧس‬.‫د‬  ‫‪ ‬ﻧﻘطﺔ اﻟﺗﻛﺎﻓؤ و ﻧﻘطﺔ ﻧﮭﺎﯾﺔ اﻟﻣﻌﺎﯾرة ‪:‬‬ ‫‪ ‬ﻧﻘطﺔ اﻟﺗﻛﺎﻓؤ ‪Equivalence point‬‬ ‫‪ ‬ھﻲ ﻧﻘطﺔ ﻧظرﯾﺔ ﯾﺻﻌب ﺗﺣدﯾدھﺎ ﺑﺷﻛل ﻋﻣﻠﻲ و ھﻲ ﺗدل ﻋﻠﻰ ﻟﺣظﺔ اﻟﺗﻔﺎﻋل اﻟﺗﺎم‬ ‫ﺑﯾن اﻟﻣﺣﻠول اﻟﻘﯾﺎﺳﻲ و اﻟﻣﺣﻠول اﻟﻣﺟﮭول ‪ ،‬أﻧﮭﺎ اﻟﻧﻘطﺔ اﻟﺗﻲ ﯾﺗﺳﺎوى ﻋﻧدھﺎ‬ ‫ﻋدد اﻟﻣﻛﺎﻓﺋﺎت اﻟﻐراﻣﯾﺔ ﻟﻠﻣﺣﻠول اﻟﻌﯾﺎري ﻣﻊ ﻋدد اﻟﻣﻛﺎﻓﺋﺎت اﻟﻐراﻣﯾﺔ ﻟﻠﻣﺣﻠول‬ ‫اﻟﻣﺟﮭول ‪.‬‬ ‫‪ ‬ﻧﻘطﺔ ﻧﮭﺎﯾﺔ اﻟﺗﻔﺎﻋل ‪End point‬‬ ‫‪ ‬ھﻲ اﻟﻧﻘطﺔ اﻟﻌﻣﻠﯾﺔ اﻟﺗطﺑﯾﻘﯾﺔ اﻟﺗﻲ ﺗﺣدد ﻟﺣظﺔ ﻧﮭﺎﯾﺔ اﻟﻣﻌﺎﯾرة ﻧﺗﯾﺟﺔ ﻟﺗﻐﯾﯾر‬ ‫ﻣﻔﺎﺟﺊ ﻓﻲ إﺣدى اﻟﺧﺻﺎﺋص اﻟﻔﯾزﯾﺎﺋﯾﺔ أو اﻟﻛﯾﻣﯾﺎﺋﯾﺔ ﻟﻠﻣﺣﻠول ﻛظﮭور ﻟون أو‬ ‫ﺗﺷﻛل راﺳب ‪ ,‬ﺗﻐﯾر ﻓﻲ ﻗﯾﻣﺔ ‪ pH‬أو اﻟﺣرارة اﻟﻧوﻋﯾﺔ أو ﺷدة اﻟﺗﯾﺎر اﻟﻛﮭرﺑﺎﺋﻲ و‬ ‫ھﻲ ﻗرﯾﺑﺔ ﻣن ﻧﻘطﺔ اﻟﺗﻛﺎﻓؤ اﻟﻧظرﯾﺔ ) ﻗﺑﻠﮭﺎ أو ﺑﻌدھﺎ (‬ ‫اﻟﻣﺷﻌر ‪Indicator :‬‬ ‫‪ ‬اﻟﻣﺷﻌر ﻋﺑﺎرة ﻋن ﻣرﻛب ﻛﯾﻣﯾﺎﺋﻲ ﺗﺗم إﺿﺎﻓﺗﮫ أﺛﻧﺎء اﻟﻣﻌﺎﯾرة ﺑﻛﻣﯾﺔ ﺿﺋﯾﻠﺔ ﺟدا‬ ‫ﺗﺗﺳــﺑب ﻓﻲ إﺣداث ﺗﻐﯾرا ﻣﻠﺣوظﺎ ﻹﺣدى اﻟﺧﺻﺎﺋص اﻟﻔﯾزﯾﺎﺋﯾﺔ أو اﻟﻛﯾﻣﯾﺎﺋﯾﺔ‬ ‫ﻟﻠﻣﺣﻠول و ﯾﺳﺎھم ﻓﻲ ﺗﺣدﯾد ﻧﻘطﺔ ﻧﮭﺎﯾﺔ اﻟﻣﻌﺎﯾرة و اﻟﺗﻲ ﯾﺟب أن ﺗﺗطﺎﺑﻖ ﻣﻊ‬ ‫ﻧﻘطﺔ اﻟﺗﻛﺎﻓؤ أو أن ﺗﻛون ﻗرﯾﺑﺔ ﺟدا ﻣﻧﮭﺎ‬ ‫د‪.‬ﻋﻼ ﯾوﻧس‬  standard ‫اﻟﻌﯾﺎري‬ Primary Standard: a high purity compound used to prepare the standard solution or to standardize the solution with. Secondary Standard: a second material used as a substitute for a suitable primary standard. This standard solution should always be standardized using a primary standard ‫ ﻋﻼ ﯾوﻧس‬.‫د‬  primary standard: ‫ﺷروط اﻟﻌﯾﺎري اﻷوﻟﻲ‬ - High purity - Stable in air and solution: composition should be unaltered in the air at ordinary or moderately high temperatures. - Not hygroscopic. - Inexpensive - Large formula weight: equivalence weight ought to be high in order to reduce the effects of small weighing error. - Readily soluble in the solvent under the given conditions of the analysis. - On titration, no interfering product(s) should be present. -Reacts rapidly and stoichiometrically with the analyte ‫ ﻋﻼ ﯾوﻧس‬.‫د‬ ‫د‪.‬ﻋﻼ ﯾوﻧس‬  ‫ﺷروط ﺗﻔﺎﻋل اﻟﻣﻌﺎﯾرة‬ For a successful titrimetric analysis, the following need to be true: The titrant should either be a standard or should be standardized The reaction should proceed to a stable and well defined equivalence point. The equivalence point must be able to be detected. The titrant’s and sample’s volume or mass must be accurately known The reaction must proceed by a definite chemistry. There should be no complicating side reactions. The reaction should be nearly complete at the equivalence point. In other words, chemical equilibrium should favour the formation of products. The reaction rate should be fast enough to be practical ‫ ﻋﻼ ﯾوﻧس‬.‫د‬ ‫ﻟﻧﺗذﻛر ﻣﺎھﻲ أﻧواع ﻣﻌﺎﯾرات اﻟﺗﺣﻠﯾل اﻟﺣﺟﻣﻲ‬ ‫ﺣﺳب اﻟﺗﻔﺎﻋل اﻟﻛﯾﻣﯾﺎﺋﻲ ﺑﯾن ‪analyte‬و‬ ‫‪titrant‬؟‬ ‫ﺗوﺟد ‪ 4‬أﻧواع‬ ‫ ‬ ‫‪ -‬ﺣﻣض أﺳﺎس ‪Acid base‬‬ ‫ ‬ ‫‪ -‬ﺗرﺳﯾب ‪Precipitation‬‬ ‫ ‬ ‫‪ -‬ﺗﻌﻘﯾد‪complexation‬‬ ‫ ‬ ‫‪ -‬أﻛﺳدة ارﺟﺎع ‪Redox‬‬ ‫ ‬ ‫د‪.‬ﻋﻼ ﯾوﻧس‬  ‫أﻧواع اﻟﻣﻌﺎﯾرة اﻟﺣﺟﻣﯾﺔ‬ ‫اﻟﻣﻌﺎﯾرة اﻟﺣﺟﻣﯾﺔ‬ ‫ﻏﯾر ﻣﺑﺎﺷرة‬ ‫ﻏﯾر ﻣﺑﺎﺷرة‬ ‫ﻣﻌﺎﯾرة ﺑﺎﻻزاﺣﺔ‬ ‫ﻣﺑﺎﺷرة‬ ‫ﻣﻌﺎﯾرة ﺑﺎﻟرﺟوع‬ ‫‪Replacement‬‬ ‫‪direct‬‬ ‫‪Back titration‬‬ ‫‪titration‬‬ ‫د‪.‬ﻋﻼ ﯾوﻧس‬  ‫أﻣﺛﻠﺔ‬ A 0.6 g sample containing sodium bicarbonate was dissolved and titrated with standard 0.11 M hydrochloric acid solution. Requiring 41 ml. The reaction is: NaHCO3 + HCl → NaCl + H2O + CO2 Calculate the percent sodium bicarbonate in the sample. ‫ ﻋﻼ ﯾوﻧس‬.‫د‬ The iron ( II ) in an acidified solution is titrated with a 0.023 M solution of potassium permanganate: 5 Fe+2 + MnO4- + 8 H+ → 5Fe+3 + Mn+2 + 4 H2O If the titration required 38 ml, how many milligrams iron are in the solution ‫ ﻋﻼ ﯾوﻧس‬.‫د‬ The sulfur content of a steel is determined by converting it to H2S gas, absorbing the H2S in 10 ml of 0.005M I2 and then the excess I2 is back titrated with 0.002 M Na2S2O3. If 2.6 ml was required for the last titration, how many milligrams sulfur are contained in the sample S→ H2S H2S + I2 → S + 2 I- + 2 H+ I2 + 2 S2O3-2 → 2 I- + S4O6-2 ‫ ﻋﻼ ﯾوﻧس‬.‫د‬  ‫اﻟﻣﺷﻌرات‬ An indicator, a substance that have distinctly different colours in acidic and basic media, is usually added to the reaction flask to signal when and if all the analytehas reacted. The use of indicators enables the end point to be observed. ‫ ﻋﻼ ﯾوﻧس‬.‫د‬ ‫د‪.‬ﻋﻼ ﯾوﻧس‬ ‫د‪.‬ﻋﻼ ﯾوﻧس‬ ‫د‪.‬ﻋﻼ ﯾوﻧس‬ ‫د‪.‬ﻋﻼ ﯾوﻧس‬  ‫ﺗﻣﺎرﯾن‬ ‫ﻛم ﻏراﻣﺎ ً ﻣن اﻟﯾود ﯾﻠزم ﻟﺗﺣﺿﯾر ‪100‬ﻣل ﻣن ﻣﺣﻠول ﺗرﻛﯾزه ‪0.1‬‬ ‫‪N‬ﻣن اﻟﯾود ؟‬ ‫ﻛم ﻋدد ﻏراﻣﺎت ﻛﻠورﯾد اﻟﺻودﯾوم ﻓﻲ اﻟﻠﺗر اﻟواﺣد ﻟﻣﺣﻠول‬ ‫ﺗرﻛﯾزه ‪0.2M‬‬ ‫ﻋﯾﻧﺔ وزﻧﮭﺎ ‪g 2.1‬ﺗﺣوي ھﯾدروﻛﺳﯾد ﺻودﯾوم وﻛرﺑوﻧﺎت ﺻودﯾوم ﻋوﯾرت‬ ‫ﺑﺣﻣض ﻛﻠور اﻟﻣﺎء ‪N 0.5‬ﻓﻠزم ‪ml 30‬ﻣن اﻟﺣﻣض ﻹﺧﺗﻔﺎء ﻟون اﻟﻔﯾﻧول ﻓﯾﺛﺎﻟﯾن‬ ‫ﺛم أﺿﯾﻔت ﻗطرﺗﺎن ﻣن ﺑرﺗﻘﺎﻟﻲ اﻟﻣﯾﺛﯾل إﻟﻰ ﻧﻔس اﻟﻣﺣﻠول وﻟزم ‪ml 5‬ﻣن اﻟﺣﻣض‬ ‫ﻟﺗﻐﯾر ﻟون اﻟﻣﺷﻌر ﻣن اﻷﺻﻔر إﻟﻰ اﻷﺣﻣر‪.‬إﺣﺳـب اﻟﻧﺳﺑﺔ اﻟﻣﺋوﯾﺔ ﻟﮭﯾدروﻛﺳﯾد‬ ‫اﻟﺻودﯾوم وﻛرﺑوﻧﺎت اﻟﺻودﯾوم ﻓﻲ اﻟﻌﯾﻧﺔ ؟‬ ‫د‪.‬ﻋﻼ ﯾوﻧس‬ Assume you are asked to prepare 500 mL of 2.5 M HCl solution, from a stock HCl solution that has a density of 1.145 g/mL and a percentage of 32% (FW of HCl = 36.5 g/mol). How many mL of the concentrated HCl solution will be needed? How many mg of Na2CO3 (FW = 106 g/mol) are needed to prepare 500 mL of a 10.00 ppm Na+ ? mg Na2CO3 = 0.10875*106 = 11.53 mg ‫اﻟحﻞ‬ ‫ ﻋﻼ ﯾوﻧس‬.‫د‬